In this puzzle, each of the items has a cost equal to the sum of the costs of its letters. Thus, the challenge is to find a value for each letter. There are a few different ways to calculate this; one path is as follows:
- PEA=$13 and PEAR=$14. Thus R=$1
- GRAPE=$21 and PEAR=$14, so G=$7
- CAR is $10, so CA=$9. Since CAT=$14, T=$5
- CARROT is $18 and CARRT adds up to $16, leaving O=$2
- TACOS is $18, and TACO adds up to $16, so S=$2
- Now that we know G and S, GAS being $13 means A=$4
- With CA=$9 and A=$4, C=$5
- Since we now know all its letters but P, PASTA=$18 gives P=$3
- Knowing P allows any of PEA/PEAR/GRAPE to give E=$6
- From TIRES, I=$6
- From TRAILER, L=$8
- From BREAD, BD=$3, which means one must be $1 and the other must be $2
- From BANANA=$25, BNN=$13, so B must be odd, thus B=$1, D=$2, and N=$6
- In CAT FOOD, F=$6
- HOT DOGS gives H=$9
- BOWL gives W=$3
- PANCAKES gives K=$7
- From HAM=$18 we get M=5
- Then JAM gives J=$7
- Now all the letters from APPLE JUICE are known except for U=$8
- Finally, KUMQUAT give Q=$3
The “Other Expenses” section references saving enough for every $6 letter. The six-dollar letters are E, F, I, and N. The only rearrangement of these which forms a common word – which is also something that might be an unexpected expense – is FINE.